Statistic confidence interval

Question

The head of maintaince at xyz rent a car believes that the mean number of miles between services is 4959. Miles. With a standard deviation of 448 miles. If he is correct what is the probability that the mean of a sample of 43 cars would be less than 5018 miles?

Answer 1

Hi Sarah-

Since the question involves the distribution of means, you need to calculate a new std deviation

std dev of the mean of 43 samples = σ/√n = 448/√43 = 68.319

Calculating a Z value = (x bar - population mean)/ std dev of mean = (5018-4959)/68.319 = 0.86

From the Z table:

P(x bar<5018) = P(Z<0.86) = .8051

I hope that helps!